Why is N^T N = I?
Since N is a unit vector, we have N^T N = N.N = |N|^2 = 1^2 = 1. I suppose 1 = I (loosely speaking) for a 1x1 matrix.
@Cake Right: $x^T y$ is just shorthand for the dot product between $x$ and $y$, since if $x$ and $y$ are column vectors we have
$$ \left[ \begin{array}{c} x_1 \ \vdots \ x_n \end{array} \right]^T \left[ \begin{array}{c} y_1 \ \vdots \ y_n \end{array} \right] = \left[ \begin{array}{ccc} x_1 & \cdots & x_n \end{array} \right] \left[ \begin{array}{c} y_1 \ \vdots \ y_n \end{array} \right] = \sum_{i=1}^n x_i y_i. $$
Shoot - the LaTeX renderer isn't working great here. In ASCII
[ x1 ]^T [ y1 ] [ y1 ] [ . ] [ . ] [ . ] [ . ] [ . ] = [ x1 ... xn ] [ . ] = x1 y1 + ... + xn yn [ . ] [ . ] [ . ] [ xn ] [ yn ] [ yn ]
Why is N^T N = I?
Since N is a unit vector, we have N^T N = N.N = |N|^2 = 1^2 = 1. I suppose 1 = I (loosely speaking) for a 1x1 matrix.
@Cake Right: $x^T y$ is just shorthand for the dot product between $x$ and $y$, since if $x$ and $y$ are column vectors we have
$$ \left[ \begin{array}{c} x_1 \ \vdots \ x_n \end{array} \right]^T \left[ \begin{array}{c} y_1 \ \vdots \ y_n \end{array} \right] = \left[ \begin{array}{ccc} x_1 & \cdots & x_n \end{array} \right] \left[ \begin{array}{c} y_1 \ \vdots \ y_n \end{array} \right] = \sum_{i=1}^n x_i y_i. $$
Shoot - the LaTeX renderer isn't working great here. In ASCII